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Which one of the following molecular geometries is correctly predicted by VSEPR theory? Consider the bond angle θ in each case.

1. NF3: central N has 3 bond pairs and 1 lone pair; bond angle θ > 109°28'
2. SF4: central S has 4 bond pairs and 2 lone pairs; bond angle θ < 90°
3. [ICl4]⁻: central I has 4 bond pairs and 2 lone pairs; bond angle θ = 90°
4. XeF4 (with one lone pair on Xe): bond angle θ = 109°28'

Correct answer: [ICl4]⁻: central I has 4 bond pairs and 2 lone pairs; bond angle θ = 90°

Solution

[ICl4]⁻ has 6 electron pairs around I (4 bonding + 2 lone), arranging in an octahedral electron geometry. The 2 lone pairs sit opposite each other (axial), giving a square planar molecular geometry with bond angles of exactly 90°, matching VSEPR prediction. All other options describe incorrect bond angles.

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